A 50 Hz, 4 Pole, 500 MVA, 22 kV turbo generator is delivering rated MVA at a power factor of 0.8. Suddenly a fault occurs reducing the electric power output by 40%. Neglect the losses and assume constant power input to the shaft. The accelerating torque in the generator at the time of fault is

Concept:

Accelerating power can be calculated as

P_a = P_m – P_e

Accelerating torque is given as

\({T_a} = \frac{{{P_a}}}{{{\omega _s}\;}}\)

Where,

P_m = Mechanical input power

P_e = Electrical output power

ω_s = Synchronous speed in radian/sec

P = Number of poles

F = Supply frequency

Synchronous speed in rpm can be calculated as

\({N_s} = \frac{{120f}}{P}\)

Also, \({\omega _s} = \frac{{2\pi {N_s}}}{{60}}\) 

Calculation:

Given-

f = 50 Hz, P = 4

Before fault

P_m = P_e = 500 cosϕ = 500 x 0.8

P_m = P_e = 400 MW

After fault P_m remains same, while P_e reduces by 40%

So electrical output at the time of fault is

P_e’ = (1 – 0.4) = 0.6 x 400

P_e’= 240 MW

Now Accelerating power at the time of fault is

P_a’ = 400 – 240 = 160 MW

Now synchronous speed is

\({N_s} = \frac{{120 \times 50}}{4} = 1500\;rpm\)

\({\omega _s} = \frac{{2\pi \times 1500}}{{60}} = 157\;radian/sec\)

Hence accelerating torque is

\({T_a} = \frac{{160 \times {{10}^6}}}{{157}} = 1.018\; \times {10^6}\;N - m\)

T_a = 1.018 MN-m